Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

• that can be solved recursively by mutual series expansion∫ψ(x) = ψ (0) (x) + d 4 yS(x − y, m)e /A(y)ψ(y)∫= ψ (0) (x) + e d 4 yS(x − y, m) /A (0) (y)ψ (0) (y)(+ e∫d 2 4 yd 4 z S(x − y, m) /A (0) (y)S(y − z, m) /A (0) (z)ψ (0) (z))− S(x − y, m)γ µ ψ (0) (y)D(y − z, m) ¯ψ(z)γ µ ψ(z) + O(e 3 ) (93)• these recursive expansions become very big very fast and a graphical representationis useful:A (0)µ (y)ψ (0) (y)A (0)µ (z)ψ (0) (z)ψ(x) ψ (0) (z)S(x − y, m)S(y − z, m)ψ (0) (z)ψ(x)S(x − y, m)D(y − z, 0)(93 ′ )• remaining open questions:– does D(x − y, m) exist?– can we compute it?• fortunately, we only need to solve(∂ 2 + m 2) D(x, m) = −δ 4 (x) (85 ′ )because (most) other propagators can be obtained by taking derivatives• since the equation is translation invariant, we should use Fourier transformation• formally (with ɛ → 0+)D(x, m) =∫ d 4 p(2π) 4 e−ipx 1p 2 − m 2 + iɛ(94)(∂ 2 + m 2) D(x, m) =∫ d 4 p(2π) 4 e−ipx −p2 + m 2p 2 − m 2 + iɛ = − ∫ d 4 p(2π) 4 e−ipx = −δ 4 (x) (95)• singularities in the integral over p 0 at ± √ |⃗p| 2 + m 2 (+iɛ is a convenient shorthandfor the choice of integration contour):17