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.


1 Answer 1

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

    Hence in the space $\mathbb{R}^{4(m+n)}$ of $m$ incoming and $n$ outgoing off-shell 4D momenta, the support of a Feynman diagram with $r$ connected components is $4(m+n-r)$-dimensional.

    Therefore for external momenta that satisfy total momentum conservation, disconnected diagrams contribute almost nowhere relatively speaking as compared to connected diagrams.

  2. Moreover, if a connected component has only 2 legs, i.e. is a connected propagator, its singularity structure is subdominant compared to the RHS of eq. (7.42). See also e.g. this related Phys.SE post.

    In particular for $2\to 2$ scattering, all the disconnected diagrams consist of 2 connected propagators, and hence have a subdominant singularity structure.

  • $\begingroup$ 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? $\endgroup$ Oct 22, 2022 at 20:47
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Oct 23, 2022 at 2:02
  • $\begingroup$ @Qmechanic Thank you very much. 1. So, your answer 1. does not say that there are no contributions from disconnected diagrams, but if we focus on regions in momentum space which do not reside "almost everywhere", we find an S-matrix corresponds to simultaneously occured events. At first I thought the word "almost everywhere" means no contribution at all (since measure may be zero), is this understanding correct? 2. Also, is your answer 2. understood in terms of residue theorem? Because poles are not order 1 if we impose the momentum conservation on disconnected diagrams in 2 -> 2 scattering. $\endgroup$
    – Keyflux
    Jun 15, 2023 at 19:53
  • $\begingroup$ Hi Keyflux, Thanks for the feedback. 1. Note that a Dirac delta distribution has support almost nowhere, but it still contributes. 2. The LSZ formula can to a large extent be understood both via real and complex analysis. $\endgroup$
    – Qmechanic
    Jun 16, 2023 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.