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As an example, using the LSZ reduction formula, the $S$ matrix element for $2\rightarrow 2$ scattering is found in Peskin and Schroeder to be

$$\langle \boldsymbol{p}_1 \boldsymbol{p}_2\rvert S \lvert \boldsymbol{k}_1 \boldsymbol{k}_2\rangle=(\sqrt{Z})^4 \sum \left(\text{Connected, amputated diagrams with incoming $k_1$, $k_2$ and outgoing $p_1$, $p_2$}\right). \tag{1}\label{1}$$

However, this clearly doesn't work in the case of identical in and out states because it neglects the disconnected contribution where the two particles do not interact. My question is therefore, where in the derivation of the LSZ reduction formula does this disconnected part fail to contribute?

I have read many times something like the LSZ reduction formula only gives the connected contribution because disconnected diagrams do not have the correct pole structure. This is true, but says nothing about the fundamental reason why it should only give the connected part. After all, we set out to relate time-ordered correlation functions to $S$ matrix elements, not to relate them to only the connected parts, and indeed, in the derivation, I see no obvious reason why it should only give the connected part.

In other words, where in the derivation do we realise that the LHS of Eq. (\ref{1}) is only the connected part?

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  • $\begingroup$ I have read many times something like 'the LSZ reduction formula only gives the connected contribution because disconnected diagrams do not have the correct pole structure'. <-- where have you read this? $\endgroup$
    – Prahar
    Apr 5, 2022 at 21:44
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    $\begingroup$ e.g. here, here, here,... $\endgroup$
    – Ghorbalchov
    Apr 5, 2022 at 21:50
  • $\begingroup$ At a cursory glance, I cannot see any of those links suggesting that LSZ theorem gives only the connected part of the $S$-matrix. All it is saying is that the connected correlator gives the connected $S$-matrix via LSZ. Similarly, the full Green's function gives the full $S$-matrix via LSZ. $\endgroup$
    – Prahar
    Apr 5, 2022 at 21:54
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    $\begingroup$ I'm not sure what to say, they explicitly write $\text{My LHS} = \text{My RHS} + \text{Disconnected Pieces}$. In other words, they are saying my LHS is only the connected part. Another example is here which says 'But $\mathcal{A}_{\text{LSZ}}$ is not the total amplitude: it is just the connected contribution!' $\endgroup$
    – Ghorbalchov
    Apr 5, 2022 at 22:01

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