3
$\begingroup$

I don't understand why in the LSZ reduction formula I need to consider only connected Feynman diagrams when I compute scattering amplitudes. From what I read in Peskin & Schroeder it seems that the disconnected diagrams contributes only to the trivial part of the $S$-matrix, but why?

The LSZ reduction formula is (consider for simplicity a complex scalar field with $m$ incoming particles and $n$ outgoing anti-particles) \begin{equation} \langle k_1...k_m; out|p_1,...,p_n; in \rangle = \left(\frac{-i}{\sqrt{Z}}\right)^{n+m} \prod_{i=1}^m \Big{[}(k_i^2 - m^2) \Big{]} \prod_{j=1}^m \Big{[}(p_j^2 - m^2) \Big{]} G(k_1,...,k_m;-p_1,...,-p_n), \end{equation} where \begin{equation} G(k_1,...,k_m;-p_1,...,-p_n) = \int d^4x_1...d^4x_md^4y_1...d^4y_n e^{i\sum_i^mk_ix_i - i \sum_j^n p_jy_j}\langle \Omega|T(\phi(x_m)...\phi(x_1)\phi^\dagger(y_1)...\phi^\dagger (y_n))|\Omega\rangle. \end{equation} Now, my problem is that this Green's function $$\langle \Omega|T(\phi(x_m)...\phi(x_1)\phi^\dagger(y_1)...\phi^\dagger (y_n))|\Omega\rangle$$ still contains disconnected feynman diagrams (not the bubble ones of course). When I decompose the $S$-matrix as $S = 1+iT$ in $1$ I can "absorb" all diagrams that have the initial state equal to the final one (i.e. no scattering has occured), but I would say that I can still have disconnected diagrams that give me final states that are different from the initial ones. In fact, I can think of a disconnected diagram which is made by two connected sub-diagrams. There are a lot of links that asked the same question, but they all seem to imply that in the previous green function there are indeed also disconnected diagrams, but they do not contibute to the connected part of the $S$-matrix. Why is only this one that counts for scattering processes? i.e. why when I solve the exercises I do not need the disconnected diagrams?

$\endgroup$
5
  • $\begingroup$ Which other links asked the same question? $\endgroup$
    – Qmechanic
    Dec 22, 2023 at 13:28
  • $\begingroup$ physics.stackexchange.com/q/638379. This one for example gives a very nice explanation, but still in the end it seems that you need to compute the connected part because the disconnected graphs can be made by smaller connected diagrams $\endgroup$
    – Alex
    Dec 22, 2023 at 13:32
  • $\begingroup$ physics.stackexchange.com/q/82387 Also this answer seems to imply the same. My problem is that I'm still very confused why I can consider only connected ones when computing the scattering amplitude. I would say disconnected diagrams can still have the right "pole structure" to give a finite result in the LSZ formula $\endgroup$
    – Alex
    Dec 22, 2023 at 13:40
  • $\begingroup$ Related: physics.stackexchange.com/q/732818/2451 , physics.stackexchange.com/q/702344/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 22, 2023 at 13:49
  • $\begingroup$ It's still not clear to me...I can still have a disconnected diagram made by two connected subdiagrams that satisfies the 4-momentum conservation (as a connected diagram). Could you please explain and expand a bit more here your answer? $\endgroup$
    – Alex
    Dec 22, 2023 at 14:03

0