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In derivation of the LSZ reduction formula in Peskin and Schroeder, on page 227, the book says

Let us analyze the relation between the diagrammatic expansion of the scalar field four-point function and the $S$-matrix element for the 2-particle $\rightarrow$ 2-particle scattering. We will consider explicitly the fully connected Feynman diagrams contributing to the correlator. By a similar analysis, it is easy to confirm that disconnected diagrams should be disregarded because they do not have the singularity structure, with a product of four poles, indicated on the right-hand side of (7.42).

$$\tag{7.42}\prod_i^n \int d^4x_i e^{ip_i\cdot x_i}\prod_1^m\int d^4y_j e^{-ik_j\cdot y_j} \langle \Omega|T\{\phi(x_1)...\phi(x_n)\phi(y_1)...\phi(y_m)\}|\Omega\rangle\thicksim \bigg(\prod_{i=1}^n\frac{\sqrt{Z}i}{p_i^2-m^2+i\epsilon}\bigg)\bigg(\prod_{j=1}^n\frac{\sqrt{Z}i}{k_j^2-m^2+i\epsilon}\bigg)\langle\boldsymbol{p}_1...\boldsymbol{p}_n|S|\boldsymbol{k}_1...\boldsymbol{k}_m\rangle.$$

My question is: how do we see that disconnected diagrams have incorrect pole structure. If a diagram is disconnected, its value would be the product of its disconnected pieces, which I think should give the correct pole structure.

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  • $\begingroup$ Those disconnected diagrams cancel out with the denominator of the cross section when doing things with the path integral formulation. It's been a while though so it's hard for me to be more specific. But you can factor them out in the numerator in a clever way and it's actually quite remarkable. $\endgroup$
    – doublefelix
    Oct 19, 2022 at 18:14

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Closely related is the fact that in 4D momentum space, a Feynman diagram/correlation function with $n$ connected components contains $n$ 4D momentum Dirac delta functions, due to spacetime translation symmetry.

Therefore for external momenta that satisfy total momentum conservation, only connected diagrams contribute almost everywhere.

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  • $\begingroup$ can we say that nonconnected diagrams imply that the sum of a subset of in momentum is a subset of the out momentum, therefore has a measure 0? $\endgroup$
    – Simplyorange
    Oct 22, 2022 at 20:47
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic ♦
    Oct 23, 2022 at 2:02

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