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Homework Help: Fourier Transform on the connected part of QFT transition prob.

  1. Apr 22, 2013 #1
    Fourier Transform on the "connected part" of QFT transition prob.

    1. The problem statement, all variables and given/known data

    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.

    3. 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.
  2. jcsd
  3. Apr 22, 2013 #2
    Have you learned how to express the green's function as a sum of feynman diagrams yet? The meaning of "connected" if that case refers to the structure of the graphs included in the expansion.

    If you aren't using diagrams, then the connected part requires you to be able to go from each external point to any other external point, by following the propagators. For example, in the very first term, you can go from x1 to x2, or x3 to x4, but not from x1 to x4. Therefore, that term is not connected.

    Scrutinizing your calculation, it looks like you forgot one term, which incidentally is the most important term in the expansion! Try to answer this question: is there a way to modify your formula D(x1,x2) so that is includes x1, x2, x3 and x4?

    (I'll help you with the other stuff, but you should be able to answer that first)
  4. Apr 22, 2013 #3
    Hello DimReg, I looked at my derivation and found a few mistakes, indeed. Thank you for the heads up. Here is my derivation | tail:

    With multiindex α := (α₁,...,αₙ) and K := 1 - iλ/24∫d⁴z(-3Δ²(z,z) + i6Δ(z,z)(∫d⁴xΔ(z,x)J(x))² + (∫d⁴xΔ(z,x)J(x))⁴), which is the interacting part due to λϕ⁴ split out of Z[J] so that Z[J] = K[J]Z₀[J] with Z₀[J] the according functional for without interaction:

    [tex]\partial_\alpha := \frac\partial{\partial J(x_{\alpha,1})} ... \frac\partial{\partial J(x_{\alpha,n})}[/tex]

    Dependence on J surpressed so that Z₀ := Z₀[J] and K := K[J], all terms which vanish at J = 0 surpressed:

    \partial_{1,2,3,4}\left[ \frac{K[J]Z_0[J]}{K[0]Z_0[0]} \right]_{J=0} = &[\partial_{1,2}Z_0\partial_{3,4}K + \partial_{3,4}Z\partial_{1,2}K +\\
    &\partial_{1,3}Z_0\partial_{2,4}K + \partial_{2,4}Z\partial_{1,3}K +\\
    &\partial_{1,4}Z_0\partial_{2,3}K + \partial_{2,3}Z\partial_{1,4}K +\\
    &K\partial_{1,2,3,4}Z + Z\partial_{1,2,3,4}K]_{J=0}

    Inserting yields, with Δᵢⱼ := Δ(xᵢ,xⱼ) and Dᵢⱼ := D(xᵢ,xⱼ):

    \partial_{1,2,3,4}\left[ K[J]Z_0[J] \right] = \frac{i}2(&\Delta_{1,2}D_{3,4} + \Delta_{3,4}D_{1,2} + \Delta_{2,4}D_{1,3} + \Delta_{1,3}D_{2,4} + \Delta_{2,4}D_{1,3} + \Delta_{1,4}D_{2,3} + \Delta_{2,3}D_{1,4}+\\
    &+\frac14 \int \mathrm d^4 z \Delta^2(z,z)(\Delta_{1,2}\Delta_{3,4}+\Delta_{1,3}\Delta_{2,4}+\Delta_{1,4}\Delta_{2,3})-\\
    &2\int \mathrm d^4 z \Delta(z,x_1) \Delta(z,x_2) \Delta(z,x_3) \Delta(z,x_4)) \lambda - (\Delta_{1,2}\Delta_{3,4}+\Delta_{1,3}\Delta_{2,4}+\Delta_{1,4}\Delta_{2,3})

    Dividing by K[0]Z[0] and to 1st order λ gives me

    \partial_{1,2,3,4}\left[ \frac{K[J]Z_0[J]}{K[0]Z_0[0]} \right]_{J=0} \approx &-(\Delta_{1,2}\Delta_{3,4}+\Delta_{1,3}\Delta_{2,4}+\Delta_{1,4}\Delta_{2,3})+\\
    &\frac{i}2(\Delta_{1,2}D_{3,4} + \Delta_{3,4}D_{1,2} + \Delta_{2,4}D_{1,3} + \Delta_{1,3}D_{2,4} + \Delta_{2,4}D_{1,3} + \Delta_{1,4}D_{2,3} + \Delta_{2,3}D_{1,4}+\\
    &\frac14 \int \mathrm d^4 z \Delta(z,z) (\Delta_{1,2}\Delta_{3,4}+\Delta_{1,3}\Delta_{2,4}+\Delta_{1,4}\Delta_{2,3} + 1) - \\
    &2\int \mathrm d^4 z \Delta(z,x_1) \Delta(z,x_2) \Delta(z,x_3) \Delta(z,x_4))\lambda

    So I still have that ∫d⁴zΔ(z,z) term in there - although as a coefficient to λ - is that okay? I guess if I had done the taylor expansion differently (expand K[J]/K[0] up to order λ and only then derive with respect to J) it would have vanished. Strange.

    But I take it -iλ∫d⁴zΔ(z,x₁)Δ(z,x₂)Δ(z,x₃)Δ(z,x₄) is what I go on with?
  5. Apr 23, 2013 #4
    Remember, an expansion to a certain order should contain every term up to that order. If you get different terms depending on the expansion you do, then maybe you aren't seeing every term. Either way though, you are a lot closer now than you were before. (It's been long enough since I've worked out these terms by hand that I've forgotten exactly what they should look like).

    Yes, do you see why that term is connected? So you should perform a fourier transform on this part to get started on the next part.
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